a. Answer The sum is A + B = [ 8 − 1 1 7 ] Explanation Given information: The matrices are, A = [ 5 − 2 3 1 ] and B = [ 3 1 − 2 6 ] Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: The sum of both matrices is, A + B = [ 5 − 2 3 1 ] + [ 3 1 − 2 6 ] A + B = [ 5 + 3 − 2 + 1 3 − 2 1 + 6 ] A + B = [ 8 − 1 1 7 ] Now, by graphing utility of matrix capability, Hence, the result is verified. b. To determine To find A − B and graphing check your result by utility. Answer The difference is A − B = [ 2 − 3 5 − 5 ] Explanation Given information: The matrices are, A = [ 5 − 2 3 1 ] and B = [ 3 1 − 2 6 ] Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: The difference of both matrices is, A − B = [ 5 − 2 3 1 ] − [ 3 1 − 2 6 ] A − B = [ 5 − 3 − 2 − 1 3 + 2 1 − 6 ] A − B = [ 2 − 3 5 − 5 ] Now, by graphing utility of matrix capability, Hence, the result is verified. c. To determine To find 3 A and graphing check your result by utility. Answer 3 A = [ 15 − 6 9 3 ] . Explanation Given information: The matrices are, A = [ 5 − 2 3 1 ] and B = [ 3 1 − 2 6 ] Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: The difference of both matrices is, 3 A = 3 [ 5 − 2 3 1 ] 3 A = [ 5 ( 3 ) − 2 ( 3 ) 3 ( 3 ) 1 ( 3 ) ] 3 A = [ 15 − 6 9 3 ] Now, by graphing utility of matrix capability, Hence, the result is verified. d. To determine To find 3 A − 2 B and graphing check your result by utility. Answer The differenceis 3 A − 2 B = [ 9 − 8 13 − 9 ] . Explanation Given information: The matrices are, A = [ 5 − 2 3 1 ] and B = [ 3 1 − 2 6 ] Concept used: Operations like addition and subtraction of matrices are only possible when they are of same dimension. Dimension of matrix with a rows and b columns is given by a × b . If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar. Calculation: The difference of both matrices is, 3 A − 2 B = 3 [ 5 − 2 3 1 ] − 2 [ 3 1 − 2 6 ] 3 A − 2 B = [ 5 ( 3 ) − 2 ( 3 ) 3 ( 3 ) 1 ( 3 ) ] − [ 3 ( 2 ) 1 ( 2 ) − 2 ( 2 ) 6 ( 2 ) ] 3 A − 2 B = [ 15 − 6 9 3 ] − [ 6 2 − 4 12 ] 3 A − 2 B = [ 15 − 6 − 6 − 2 9 − ( − 4 ) 3 − 12 ] 3 A − 2 B = [ 9 − 8 9 + 4 − 9 ] 3 A − 2 B = [ 9 − 8 13 − 9 ] Now, by graphing utility of matrix capability, Hence, the result is verified.

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

See similar textbooks

Videos

Question

Chapter 7.5, Problem 13E

To determine

To find A+B and graphing check your result by utility.

Expert Solution

Answer to Problem 13E

The sum is A+B=[8−117]

Explanation of Solution

Given information:

The matrices are,

A=[5−231] and B=[31−26]

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The sum of both matrices is,

A+B=[5−231]+[31−26]A+B=[5+3−2+13−21+6]A+B=[8−117]

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 13E , additional homework tip 1

Hence, the result is verified.

To determine

To find A−B and graphing check your result by utility.

Expert Solution

Answer to Problem 13E

The difference is A−B=[2−35−5]

Explanation of Solution

Given information:

The matrices are,

A=[5−231] and B=[31−26]

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The difference of both matrices is,

A−B=[5−231]−[31−26]A−B=[5−3−2−13+21−6]A−B=[2−35−5]

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 13E , additional homework tip 2

Hence, the result is verified.

To determine

To find 3A and graphing check your result by utility.

Expert Solution

Answer to Problem 13E

3A=[15−693] .

Explanation of Solution

Given information:

The matrices are,

A=[5−231] and B=[31−26]

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The difference of both matrices is,

3A=3[5−231]3A=[5(3)−2(3)3(3)1(3)]3A=[15−693]

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 13E , additional homework tip 3

Hence, the result is verified.

To determine

To find 3A−2B and graphing check your result by utility.

Expert Solution

Answer to Problem 13E

The differenceis 3A−2B=[9−813−9] .

Explanation of Solution

Given information:

The matrices are,

A=[5−231] and B=[31−26]

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by a×b .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The difference of both matrices is,

3A−2B=3[5−231]−2[31−26]3A−2B=[5(3)−2(3)3(3)1(3)]−[3(2)1(2)−2(2)6(2)]3A−2B=[15−693]−[62−412]3A−2B=[15−6−6−29−(−4)3−12]3A−2B=[9−89+4−9]3A−2B=[9−813−9]

Now, by graphing utility of matrix capability,

Precalculus with Limits: A Graphing Approach, Chapter 7.5, Problem 13E , additional homework tip 4

Hence, the result is verified.

Chapter 7.5, Problem 12E

Chapter 7.5, Problem 14E

Chapter 7 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 7.1 - Prob. 1E Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.